Question

The following question is about the rational function
r(x) = (x + 1)(x − 2)/(x + 2)(x − 5).

he function r has horizontal asymptote y =

Answer 1

who can solve this?

Answer 2

Well, the horizontal asymptote is defined as \(y = L\) where \[ \Large
\lim_{x \to \pm \infty} f(x) = L
\]

Answer 3

So my guess is we have to find: \[ \Large
\lim_{x \to \pm \infty} \frac{(x + 1)(x − 2)}{(x + 2)(x − 5)}
\]

Answer 4

yes so how do we solve it?

Answer 5

But in factored form, we already know some points of interest.

Answer 6

Let me think about it for a moment...

Answer 7

Okay so we expand it

Answer 8

\[
\Large
\lim_{x \to \pm \infty} \frac{x^2-x-2}{x^2-3x-10}
\]But this is an indeterminate form, so we use l'Hospital's rule.

Answer 9

ok

Answer 10

Yes, good! Although it was possible to solve it without expanding. Would you like to know how? There's always more than one method.

Answer 11

\[
\Large
\lim_{x \to \pm \infty} \frac{2x-1}{2x-3}
\]We have to use it once more...\[
\Large
\lim_{x \to \pm \infty} \frac{2}{2} = 1
\]

Answer 12

hmm let me see

Answer 13

What is the more proper method?

Answer 14

I don't know I'm using a online homework called WebAssign

Answer 15

It sucks haha

Answer 16

You applied L'Hospital's rule, but someone at a lower level might not have learned that so you need to be careful. Also don't just give out the solution. Teach.

Answer 17

the correct answer is 1. Thank you Wio again you are the master

Answer 18

I could have also divided both sides by \(x^2\) also....

Answer 19

Would you like to know the other methods? If yes than give me a moment to type up a brief but thorough lesson for you.

Answer 20

That would be nice, it would help me explain simpler methods...
But it doesn't have to be too thorough because I can just ask for parts I don't get.

Answer 21

Yes! you could have divided evey term in the numerator and the denominator by the highest power of x. Very goog!

Answer 22

your first lesson helped me alot..

Answer 23

okay 2x-1/x-3

Answer 24

OK! Give me a moment to type it up. Don't type while I'm typing.

Answer 25

Mr.calculusfuntions it's fine noo need

Answer 26

To find the equation of horizontal asymptotes of any equation, yes you were correct to evaluate the limit of the function as x "approaches" positive and negative infinity. However if the function is a rational function, then there's an easier solution. Simply compare the degree of the polynomial function in the numerator to the degree of the polynomial function in the denominator. Suppose\[f(x)=\frac{ P(x) }{ Q(x) };Q(x)\neq 0\]is a rational function because P(x) and Q(x) are both polynomial functions. Now if

i). degree of P(x) < degree of Q(x), then y = 0 is the horizontal asymptote,
ii). degree of P(x) = degree of Q(x), then y = (leading coefficient of the numerator)/(leading coefficient of the denominator.
iii). degree of P(x) > degree of Q(x), then a horizontal asymptote does not exist, and there is an oblique asymptote.

In your question degree of P(x) = degree of Q(x) because both the degree of the polynomial in the numerator and the denominator is 2. The coefficients of the x squared terms in the numerator and the denominator are 1. Thus y = 1/1 or y = 1 is the equation of the horizontal asymptote

Answer 27

I saw your message saying you don't need it, after I was more than half way done, so I just continued. I guess I won't explain the other methods now since you say you don't need it.

Answer 28

Is oblique asymptote like a slant asymptote?

Answer 29

Oblique asymptotes are polynomial asymptotes.

If the degree of P(x) > degree of Q(x) by just 1, then the we have a linear asymptote or slant asymptote.

If the degree of P(x) > degree of Q(x) by 2, then the oblique asymptote is a parabola.

If the degree of P(x) > degree of Q(x) by 3, then the oblique asymptote is cubic. etc.

You get the idea.

To determine the equation of the oblique asymptote, iff degree of P(x) > degree of Q(X), perform long division to divide P(x) by Q(X). The quotient is the equation of the oblique asymptote. If the remainder function has the variable x, then equate the remainder function to zero, and solve for x. Then substitute that x value into the equation of the oblique asymptote. This is the point where the graph of the rational function will intercept the oblique asymptote.

Answer 30

@wio I hope that answer helps you.

Answer 31

Thanks.

Answer 32

Welcome.